Demystifying Apparent Wind - Part 1
A very common problem my students have voiced is that they have trouble figuring out where the wind's coming from. On the one hand, it shouldn't be so hard - turn your head and feel the wind on your ears. When it's the same on both ears, you're either looking straight into the wind or straight away from it. Or (less accurate) look at the yarn on the side stay.
Yet it's not so simple. You think of the wind as coming from the same direction, but when you sail, it doesn't seem to do that. It seems to shift a lot. And "you told me we couldn't sail higher than about 45 degrees into the wind, but that yarn on the stay has a much smaller angle than that". And (yet more observent) "We're sideways to the waves, so we should be on a beam reach, but the yarn says we're still somewhat upwind".
Welcome to the world of apparent wind. This is a simple concept to explain and understand at a high level, but very hard to get at a detailed level. Everyone gets the simple explanation - you're peddling a bike at 10 knots on a calm day, what do you feel? A 10 knot headwind. The speed you're generating adds to the wind speed to create the wind you feel, the apparent wind. The apparent wind is what you and the bike feel. Peddaling 10 knots in a 10 knot headwind, and you're pumping against a 20 knot apparent wind. Doing the same in a 10 knot tail wind, and Bob's very much your uncle.
Those with a math background easily grasp that this is a vector algebra problem - the boat wind speed adds up with the true wind speed as vectors, where both the speeds and the directions interact. But even if you get that, it's truly hard to see how it all plays out on the water. And I'm speaking as someone with a graduate education in mathematics. In editing this blog, I realized that I had messed up a calculation in my first draft. If you sit down to do the calculations, you have to determine what your boat speed will be at a given true wind speed and point of sail (angle of the boat to the wind). Polar performance diagrams will show this, but good luck finding these for any of our dinghy's.
So my point is that it's quite difficult to build a mental model of this. Instead, you can get a feel of how it happens on the water without trying to understand why, exactly. I did the math from a guess of a polar diagram for a boat similar to a Laser Bahia at 5 knots wind. If you're not racing, it doesn't matter how accurate this is. But it should be pretty typical.
Remember that the wind the boat feels (and you feel) is not the true wind, it's the apparent wind. So you're trimming the sails for the apparent wind, and the true wind is in practical terms irrelevant.
I'll work through some examples at various points of sail. Here's what it looks like close hauled:
You're sailing 45 degrees to the true wind, but you're sailing about 30 degrees to the apparent wind. Look at the yarn on the side stay, and you'll see that angle. Remember it, as it's one guide to show you that you're going as high as you can into the wind (a better guide is the jib tell-tales when the jib is sheeted in hard, but unless you're sitting forward in the boat, you may not be able to see them - in that case, a good crew can really help). It feels really fast, as your boat speed is pretty much addingdirectly to the true wind speed to produce apparent wind speed. If you're trimmed correctly, your main sail is sheeted in hard, so the lift force from the sail (perpendicular to the sail) is more or less sideways to the boat, causing a fair amount of heeling force, which you have to balance with your weight.
Now let's go to beam reach:
Beam reach is when the apparent wind is 90-degrees (sideways) to the boat. Because of the way boat speed and true wind interact, you've got to go downwind 130 degrees to the true wind (40 degrees past perpendicular) to get to a beam reach. Beam reach is the fastest (non-planing) point of sail. The sail is still working like a wing (more power than the downwind parachute mode), but the properly trimmed sails are out farther, so the lift force (always perpendicular to the sail) is more aligned with the direction the boat is going. More propulsion, less heel. It doesn't feel as fast as going close-hauled, as the apparent wind speed is less. But the boat speed is faster.
Now let's go downwind to a dead run:
As we turn through a broad reach and get the wind behind us for the first time, everything seems to slow down. That's because when the wind is behind us, our boat speed cancels some of the true wind, so the apparent wind speed slows down. Note that here it's less than half of the apparent wind close-hauled with the same true wind. If you're racing, you'd better be prepared for a big increase in apparent wind speed as you round the leeward mark.
Here's an animation that puts all of this together (click on it to see it):
How much does any of this matter? If you're an experienced sailor, you just know it. You're tuned to the apparent wind, and you're setting your boat for that. So when you fly the kite, the apparent wind moves way forward, and you deal with that. But if you're just learning sailing, it does matter, as it explains the confusing things you're seeing on the water. It gives you a visual model of sail trim.
The yarn on the shrouds will also tell you the precise apparent wind direction, in anything over a few knots.
It would be a great senior project for someone to make some polars for our dinghies! There are enough people in the club with GPS trackers and portable anemometers that this should be feasible.
That didn't work, try here....
This is what I used for the hypothetical dinghy in the blog:
Hi John. Great writeup, thanks for sharing. One question, you wrote:
"Beam reach is when the apparent wind is 90-degrees (sideways) to the boat. Because of the way boat speed and true wind interact, you've got to go downwind 130 degrees to the true wind (40 degrees past perpendicular) to get to a beam reach."
All the diagrams I've seen define beam reach as 90 degrees to the true wind, not the apparent wind. I thought that if you're at 90 degrees to the true wind, no matter how fast you go, you're still on a beam reach... you don't have to keep going down to stay on a beam. Isn't that correct?
Yes, I use a diagram in "ground school" to teach points of sail, with a wind circle (true wind) and a model sailboat. So I guess I teach that a beam reach is 90 degrees to the true wind, but I do that because to do otherwise would be too confusing. And students don't know enough to make the true/apparent distinction.
Let me put it a different way. If you're sailing, how do you know you're on a beam reach? You can look at the woolen tell-tale on the side-stay and see it at 90-degrees to the boat, or you can feel the wind equally on both ears as you look perpendicular to the boat, or whatever. What you're looking at is the apparent wind. Next time you're out in any visible waves, note the angle of the boat to the waves when you're on a beam reach (the waves are approximately coming from the true wind direction), and you'll see that on a beam reach you're more than 90-degrees to the true wind.
You and the boat feel apparent wind only. It's not easy to tell where the true wind is coming from. Let's assume it's constant direction and velocity (never!). In the moment, there's no way you can determine it (other than looking at a windsock or a smokestack - there's nothing on the boat that indicates it). At best, you can tack close-hauled to close-hauled, noting the course on each tack and splitting the difference.
I hope this makes sense.
Indeed, more efficient than tack and split the difference. But it still shows that there's nothing you can look at in the moment to see the true wind (the wave direction is an approximate average historical, but not instantaneous).
Great article. It can also be trippy if you're sailing a catamaran or skiff and your apparent wind lets you sail faster than the actual wind speed!
If the wind is 20 knots and the boat speed is 10 knots and we're on an apparent beam reach, how far off the true wind are we?
It's a (relatively) simple trigonometry problem. The answer is 120 degrees.
Here's the math: